Let the lines L _ 1 : (x + 5)/(3) = (y + 4)/(1) = (z - α)/(- 2) and...

Given,

The lines ${L}_{1}: \dfrac{x + 5}{3} = \dfrac{y + 4}{1} = \dfrac{z - \alpha}{- 2}$ and ${L}_{2}: 3x + 2y + z - 2 = 0 = x - 3y + 2z - 13$ be coplanar,

Now direction ratio of line ${L}_{2}$ will be,

\overset{\to}{n}2 = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & k \\ 3 & 2 & 1 \\ 1 & - 3 & 2\end{cases}\right|= \overset{^}{i}\left(7\right) - \overset{^}{j}\left(5\right) + k\left(- 11\right)

Let $R$ lies on ${L}_{2}$ and let $z = 0$

$∴ 3x + 2y = 2$ and $x - 3y = 13$

Now solving above two equation we get,

$11y = – 39 + 2 = – 37$$⇒ y = \dfrac{- 37}{11}$

And $x = 13 + 3y$$⇒ x = 13 - \dfrac{37 \times 3}{11}$

$⇒ x = \dfrac{143 - 111}{11} = \dfrac{32}{11}$

Hence, the point $R\left(\dfrac{32}{11}, \dfrac{- 37}{11}, 0\right)$

Also ${L}_{1}$, ${L}_{2}$ and given point on line ${L}_{1}$ to $R$ are coplanar,

So, by coplanar condition we get,\left|$\begin{cases}- 5 - \dfrac{32}{11} & - 4 + \dfrac{37}{11} & \alpha \\ 3 & 1 & - 2 \\ 7 & - 5 & - 11\end{cases}$\right| = 0$</p><p>$⇒ \left|$\begin{cases}\dfrac{- 87}{11} & \dfrac{- 7}{11} & \alpha \\ 3 & 1 & - 2 \\ 7 & - 5 & - 11\end{cases}$\right| = 0$</p><p>$⇒ \alpha = 7$</p><p>Now let,$P ( 3k - 5, k - 4, – 2k + 7 )$is nearest to point$( – 4, – 3, 2 )$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/d51380dc-d716-445f-badc-b98f33484c64.png" style="width: 300px; height: 221px;" /></p><p>So, by perpendicular condition we get,</p><p>$\overset{\to}{QP} . \overset{\to}{n}1 = 0$</p><p>$⇒ \left(\left(3k - 1\right)\overset{^}{i} + \left(k - 1\right)\overset{^}{j} + \left(- 2k + 5\right)k\right) \cdot \left(3\overset{^}{i} + \overset{^}{j} - 2k\right) = 0$</p><p>$⇒ 9k – 3 + k – 1 + 4k – 10 = 0$</p><p>$⇒ 14k = 14 ⇒ k = 1$</p><p>$∴ P\left(- 2, - 3, 5\right) ≡ \left(a, b, c\right)$</p><p>Hence, \left|a\right| + \left|b\right| + \left|c\right| = 2 + 3 + 5 = 10$